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My text gave the following statement for the FTOC:

Let $f$ be an integrable function on $[a, b]$, $F$ be the antiderivative of $f$. Then $$ \int_a^b f \, du = F(b) - F(a)$$ Why does the book need to specify that $F$ is an antiderivative of $f$?
Is it suggesting that while $f$ is integrable, the derivative of the integral may not be $f$?
I don't recall encountering such a situation in Calculus .... TY

Michael Hardy
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Andy Tam
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    Well, because $,F,$ must be an antiderivative, or if you prefer a primitive function, of $,f,$...To talk about "the derivative of the integral" is, imo, a little confusing: the derivative wrt what of what integral? – DonAntonio Jun 15 '13 at 18:43
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    Does it say that $F$ is the antiderivative, or does it say that $F$ is an antiderivative? The latter is correct, and is required because if $F$ is an antiderivative, so is $F+c$ for any constant $c$. – Brian M. Scott Jun 15 '13 at 18:44

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Just because a function is integrable, doesn't mean it has an antiderivative. Consider, for example, the function $f$ such that $f(0) = 1$ and $f(x) = 0$ for all $x \ne 0$.

In this case, the derivative of $\int_c^x f(t) dt$ is not equal to $f(x)$ at $x=0$.

Yoni Rozenshein
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Your misconception is typical: let us see if the following information can help you.
I don't use the symbol $[a,b]$ to simplify.

The point is to use the function is a derivative instead of the function has an antiderivative.

Not every derivative is Riemann integrable but (the FTC)

if a derivative $F'$ is Riemann integrable, then $$\int_a^b F'(x)\,dx=F(b)-F(a)$$

As for your calculus memories, there you meet only continuous derivatives and since every continuous function is Riemann integrable $\ldots$

Addendum: if you suspect vicious circles, note that Riemann integration is not the only way to prove that a function is a derivative (i.e. has an antiderivative): see my question.

Community
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Tony Piccolo
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